Question

Prove that the derivative of an odd function is always an even function.

Answer 1

Since f is differentiable, so by definition we have
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }(-x)=\underset{h\to 0}{lim}\frac{f(-x+h)-f(-x)}{h}=\underset{h\to 0}{lim}\frac{f(-(x-h))-f(-x)}{h}$

If $f\left(x\right)$ is odd, then $f(-x)=-f\left(x\right)$

$\therefore f^{\prime }(-x)=\underset{h\to 0}{lim}\frac{-f(x-h)+f\left(x\right)}{h}=-\underset{h\to 0}{lim}\frac{f(x-h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f(x-h)-f\left(x\right)}{-h}=f^{\prime }\left(x\right)$

Hence, $f^{\prime }(-x)=f^{\prime }\left(x\right)$

Hence, derivative of an odd function is even